摘要： 如图.在平面直角坐标系中.抛物线=-++经过A.B(.0). C(.0)三点.且-=5． (1)求.的值, (2)在抛物线上求一点D.使得四边形BDCE是以BC为对 角线的菱形, (3)在抛物线上是否存在一点P.使得四边形BPOH是以OB为对角线的菱形?若存在.求出点P的坐标.并判断这个菱形是否为正方形?若不存在.请说明理由． 解:(1)解法一: ∵抛物线=-++经过点A. ∴=-4 --1分 又由题意可知..是方程-++=0的两个根. ∴+=. =-=6··································································· 2分 由已知得(-)=25 又(-)=(+)-4=-24 ∴ -24=25 解得=± ··········································································································· 3分 当=时.抛物线与轴的交点在轴的正半轴上.不合题意.舍去． ∴=-． ·········································································································· 4分 解法二:∵.是方程-++c=0的两个根. 即方程2-3+12=0的两个根． ∴=.··········································································· 2分 ∴-==5. 解得 =±······························································································· 3分 (2)∵四边形BDCE是以BC为对角线的菱形.根据菱形的性质.点D必在抛物线的对称轴上. 5分 又∵=---4=-(+)+ ································· 6分 ∴抛物线的顶点(-.)即为所求的点D．······································· 7分 (3)∵四边形BPOH是以OB为对角线的菱形.点B的坐标为. 根据菱形的性质.点P必是直线=-3与 抛物线=---4的交点. ···························································· 8分 ∴当=-3时.=-×(-3)-×(-3)-4=4. ∴在抛物线上存在一点P.使得四边形BPOH为菱形． ·················· 9分 四边形BPOH不能成为正方形.因为如果四边形BPOH为正方形.点P的坐标只能是.但这一点不在抛物线上．······································································································· 10分

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